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5.5: Dividing Polynomials

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Learning Objectives

By the end of this section, you will be able to:

  • Divide monomials
  • Divide a polynomial by a monomial
  • Divide polynomials using long division
  • Divide polynomials using synthetic division
  • Divide polynomial functions
  • Use the remainder and factor theorems

Before you get started, take this readiness quiz.

  1. Add: 3d+xd.
    If you missed this problem, review [link].
  2. Simplify: 30xy35xy.
    If you missed this problem, review [link].
  3. Combine like terms: 8a2+12a+1+3a25a+4.
    If you missed this problem, review [link].

Dividing Monomials

We are now familiar with all the properties of exponents and used them to multiply polynomials. Next, we’ll use these properties to divide monomials and polynomials.

Example 5.5.1

Find the quotient: 54a2b3÷(6ab5).

Solution

When we divide monomials with more than one variable, we write one fraction for each variable.

54a2b3÷(6ab5)Rewrite as a fraction.54a2b36ab5Use fraction multiplication.546·a2a·b3b5Simplify and use the Quotient Property.9·a·1b2Multiply.9ab2

Try It! 5.5.1

Find the quotient: 72a7b3÷(8a12b4).

Answer

9a5b

Try It! 5.5.2

Find the quotient: 63c8d3÷(7c12d2).

Answer

9dc4

Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.

Example 5.5.2

Find the quotient: 14x7y1221x11y6.

Solution

Be very careful to simplify 1421 by dividing out a common factor, and to simplify the variables by subtracting their exponents.

14x7y1221x11y6Simplify and use the Quotient Property.2y63x4

Try It! 5.5.3

Find the quotient: 28x5y1449x9y12.

Answer

4y27x4

Try It! 5.5.4

Find the quotient: 30m5n1148m10n14.

Answer

58m5n3

Divide a Polynomial by a Monomial

Now that we know how to divide a monomial by a monomial, the next procedure is to divide a polynomial of two or more terms by a monomial. The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition. The sum y5+25 simplifies to y+25. Now we will do this in reverse to split a single fraction into separate fractions. For example, y+25 can be written y5+25.

This is the “reverse” of fraction addition and it states that if a, b, and c are numbers where c0, then a+bc=ac+bc. We will use this to divide polynomials by monomials.

definition: DIVISION OF A POLYNOMIAL BY A MONOMIAL

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Example 5.5.3

Find the quotient: (18x3y36xy2)÷(3xy).

Solution

(18x3y36xy2)÷(3xy)Rewrite as a fraction.18x3y36xy23xyDivide each term by the divisor. Be careful with the signs!18x3y3xy36xy23xySimplify.6x2+12y

Try It! 5.5.5

Find the quotient: (32a2b16ab2)÷(8ab).

Answer

4a+2b

Try It! 5.5.6

Find the quotient: (48a8b436a6b5)÷(6a3b3).

Answer

8a5b+6a3b2

Divide Polynomials Using Long Division

Divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.

This figure shows the long division of 875 divided by 25. 875 is labeled dividend and 25 is labeled divisor. The result of 35 is labeled quotient. The 3 in 35 is determined from the number of times we can divide 25 into 87. Multiplying 25 and 3 results in 75. 75 is subtracted from 87 to get 12. The 5 from 875 is dropped down to make 12 into 125. The 5 in 35 is determined from the number of times was can divide 25 into 125. Since 25 goes into 125 evenly there is no remainder. The result of subtracting 125 from 125 is 0 which is labeled remainder.

We check division by multiplying the quotient by the divisor. If we did the division correctly, the product should equal the dividend.

35·25875

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

Example 5.5.4

Find the quotient: (x2+9x+20)÷(x+5).

Solution

(x2+9x+20)÷(x+5)
Write it as a long division problem.
Be sure the dividend is in standard form.

x+5x2+9x+200

Divide x2 by x. It may help to ask yourself, “What do I need
to multiply x by to get x2?”

xx+5x2+9x+200

Put the answer, x, in the quotient over the x term.
Multiply x times x+5. Line up the like terms under the dividend.

xx+5x2+9x+200x2+5x_

Subtract x2+5x from x2+9x.
You may find it easier to change the signs and then add.
Then bring down the last term, 20.

xx+5x2+009x+200x2+(5x)_4x+20


Divide 4x by x. It may help to ask yourself, “What do I
need to multiply x by to get 4x?”
Put the answer, 4, in the quotient over the constant term.

x+04x+5x2+009x+200x2+(5x)_4x+20

Multiply 4 times x+5.

x+04x+5x2+009x+200x2+(5x)_4x+204x+20_

Subtract 4x+20 from 4x+20.

x+04x+5x2+009x+200x2+(5x)_4x+204x+(20)_0

Check:

Multiply the quotient by the divisor.(x+4)(x+5)You should get the dividend.x2+9x+20

 
Try It! 5.5.7

Find the quotient: (y2+10y+21)÷(y+3).

Answer

y+7

Try It! 5.5.8

Find the quotient: (m2+9m+20)÷(m+4).

Answer

m+5

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In the next example, we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Look back at the dividends in previous examples. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in this example will be x4x2+5x6. It is missing an x3 term. We will add in 0x3 as a placeholder.

Example 5.5.5

Find the quotient: (x4x2+5x6)÷(x+2).

Solution

Notice that there is no x3 term in the dividend. We will add 0x3 as a placeholder.

  .
Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. .
Divide x4 by x.
Put the answer, x3, in the quotient over the x3 term.
Multiply x3 times x+2. Line up the like terms.
Subtract and then bring down the next term.
.
Divide 2x3 by x.
Put the answer, 2x2, in the quotient over the x2 term.
Multiply 2x2 times x+1. Line up the like terms
Subtract and bring down the next term.
.
Divide 3x2 by x.
Put the answer, 3x, in the quotient over the x term.
Multiply 3x times x+1. Line up the like terms.
Subtract and bring down the next term.
.
Divide x by x.
Put the answer, 1, in the quotient over the constant term.
Multiply 1 times x+1. Line up the like terms.
Change the signs, add.

Write the remainder as a fraction with the divisor as the denominator.
.
To check, multiply (x+2)(x32x2+3x14x+2).
The result should be x4x2+5x6.
 
Try It! 5.5.9

Find the quotient: (x47x2+7x+6)÷(x+3).

Answer

x33x2+2x+1+3x+3

Try It! 5.5.10

Find the quotient: (x411x27x6)÷(x+3).

Answer

x33x22x13x+3

In the next example, we will divide by 2a3. As we divide, we will have to consider the constants as well as the variables.

Example 5.5.6

Find the quotient: (8a3+27)÷(2a+3).

Solution

This time we will show the division all in one step. We need to add two placeholders in order to divide.

  .
  .

To check, multiply (2a+3)(4a26a+9).

The result should be 8a3+27.

Try It! 5.5.11

Find the quotient: (x364)÷(x4).

Answer

x2+4x+16

Try It! 5.5.12

Find the quotient: (125x38)÷(5x2).

Answer

25x2+10x+4

Divide Polynomials using Synthetic Division

As we have mentioned before, mathematicians like to find patterns to make their work easier. Since long division can be tedious, let’s look back at the long division we did in Example and look for some patterns. We will use this as a basis for what is called synthetic division. The same problem in the synthetic division format is shown next.

The figure shows the long division of 1 x squared plus 9 x plus 20 divided by x plus 5 right next to the same problem done with synthetic division. In the long division problem, the coefficients of the dividend are 1 and 9 and 20 and the zero of the divisor is negative 5. In the synthetic division problem, we just write the numbers negative 5 1 9 20 with a line separating the negative 5. In the long division problem, the subtracted terms are 5 x and 20. In the synthetic division problem the second line is the numbers negative 5 and negative 20. The remainder of the problem is 0 and the quotient is x plus 4. The synthetic division puts these coefficients as the last line 1 4 0.

Synthetic division basically just removes unnecessary repeated variables and numbers. Here all the x and x2 are removed. as well as the x2 and 4x as they are opposite the term above.

  • The first row of the synthetic division is the coefficients of the dividend. The 5 is the opposite of the 5 in the divisor.
  • The second row of the synthetic division are the numbers shown in red in the division problem.
  • The third row of the synthetic division are the numbers shown in blue in the division problem.

Notice the quotient and remainder are shown in the third row.

Synthetic division only works when the divisor is of the form xc.

The following example will explain the process.

Example 5.5.7

Use synthetic division to find the quotient and remainder when 2x3+3x2+x+8 is divided by x+2.

Solution

Write the dividend with decreasing powers of x. .
Write the coefficients of the terms as the first
row of the synthetic division.
.
Write the divisor as xc and place c
in the synthetic division in the divisor box.
.
Bring down the first coefficient to the third row. .
Multiply that coefficient by the divisor and place the
result in the second row under the second coefficient.
.
Add the second column, putting the result in the third row. .
Multiply that result by the divisor and place the
result in the second row under the third coefficient.
.
Add the third column, putting the result in the third row. .
Multiply that result by the divisor and place the
result in the third row under the third coefficient.
.
Add the final column, putting the result in the third row. .
The quotient is 2x21x+3 and the remainder is 2.  

The division is complete. The numbers in the third row give us the result. The 2   1   3 are the coefficients of the quotient. The quotient is 2x21x+3. The 2 in the box in the third row is the remainder.

Check:

(quotient)(divisor)+remainder=dividend(2x21x+3)(x+2)+2?=2x3+3x2+x+82x3x2+3x+4x22x+6+2?=2x3+3x2+x+82x3+3x2+x+8=2x3+3x2+x+8

Try It! 5.5.13

Use synthetic division to find the quotient and remainder when 3x3+10x2+6x2 is divided by x+2.

Answer

3x2+4x2; 2

Try It! 5.5.14

Use synthetic division to find the quotient and remainder when 4x3+5x25x+3 is divided by x+2.

Answer

4x23x+1;1

In the next example, we will do all the steps together.

Example 5.5.8

Use synthetic division to find the quotient and remainder when x416x2+3x+12 is divided by x+4.

Solution

The polynomial x416x2+3x+12 has its term in order with descending degree but we notice there is no x3 term. We will add a 0 as a placeholder for the x3 term. In xc form, the divisor is x(4).

The figure shows the results of using synthetic division with the example of the polynomial x to the fourth power minus 16 x squared plus 3 x plus 12 divided by x plus 4. The divisor number if negative 4. The first row is 1 0 negative 16 3 12. The first column is 1 blank 1. The second column is negative 16 16 0. The third column is 3 0 3. The fourth column is 12 negative 12 0.

We divided a 4th degree polynomial by a 1st degree polynomial so the quotient will be a 3rd degree polynomial.

Reading from the third row, the quotient has the coefficients 1   4   0   3, which is x34x2+3. The remainder
is 0.

Try It! 5.5.15

Use synthetic division to find the quotient and remainder when x416x2+5x+20 is divided by x+4.

Answer

x34x2+5; 0

Try It! 5.5.16

Use synthetic division to find the quotient and remainder when x49x2+2x+6 is divided by x+3.

Answer

x33x2+2; 0

Divide Polynomial Functions

Just as polynomials can be divided, polynomial functions can also be divided.

definition: DIVISION OF POLYNOMIAL FUNCTIONS

For functions f(x) and g(x), where g(x)0,

(fg)(x)=f(x)g(x)

Example 5.5.9

For functions f(x)=x25x14 and g(x)=x+2, find:

  1. (fg)(x)
  2. (fg)(4).

Solution

Equation shows f over g of x equals f of x divided by g of x. This is translated into a division problem showing x squared minus 5x minus 14 divided by x plus 2. The quotient is x minus 7.

Substitute for f(x) and g(x).(fg)(x)=x25x14x+2Divide the polynomials.(fg)(x)=x7

ⓑ In part ⓐ we found (fg)(x) and now are asked to find (fg)(4).

(fg)(x)=x7To find (fg)(4), substitute x=4.(fg)(4)=47(fg)(4)=11

Try It! 5.5.17

For functions f(x)=x25x24 and g(x)=x+3, find:

  1. (fg)(x)
  2. (fg)(3).
Answer a

(fg)(x)=x8

Answer b

(fg)(3)=11

Try It! 5.5.18

For functions f(x)=x25x36 and g(x)=x+4, find:

  1. (fg)(x)
  2. (fg)(5).
Answer a

(fg)(x)=x9

Answer b

(fg)(x)=x9

Use the Remainder and Factor Theorem

Let’s look at the division problems we have just worked that ended up with a remainder. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as xc, the value of the function at c, f(c), is the same as the remainder from the division problem.

Dividend Divisor xc Remainder Function f(c)
x4x2+5x6 x(2) 4 f(x)=x4x2+5x6 4
3x32x210x+8 x2 4 f(x)=3x32x210x+8 4
x416x2+3x+15 x(4) 3 f(x)=x416x2+3x+15 3

To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and add the remainder. In function notation we could say, to get the dividend f(x), we multiply the quotient, q(x) times the divisor, xc, and add the remainder, r.

  .
If we evaluate this at c, we get: .
  .
  .

This leads us to the Remainder Theorem.

Definition: REMAINDER THEOREM

If the polynomial function f(x) is divided by xc, then the remainder is f(c).

Example 5.5.10

Use the Remainder Theorem to find the remainder when f(x)=x3+3x+19 is divided by x+2.

Solution

To use the Remainder Theorem, we must use the divisor in the xc form. We can write the divisor x+2 as x(2). So, our c is 2.

To find the remainder, we evaluate f(c) which is f(2).

  .
To evaluate f(2), substitute x=2. .
Simplify. .
  .
  The remainder is 5 when f(x)=x3+3x+19 is divided by x+2.
Check:
Use synthetic division to check.
 
.  
The remainder is 5.  
Try It! 5.5.19

Use the Remainder Theorem to find the remainder when f(x)=x3+4x+15 is divided by x+2.

Answer

1

Try It! 5.5.20

Use the Remainder Theorem to find the remainder when f(x)=x37x+12 is divided by x+3.

Answer

6

When we divided 8a3+27 by 2a+3 in Example the result was 4a26a+9. To check our work, we multiply 4a26a+9 by 2a+3 to get 8a3+27.

(4a26a+9)(2a+3)=8a3+27

Written this way, we can see that 4a26a+9 and 2a+3 are factors of 8a3+27. When we did the division, the remainder was zero.

Whenever a divisor, xc, divides a polynomial function, f(x), and resulting in a remainder of zero, we say xc is a factor of f(x).

The reverse is also true. If xc is a factor of f(x) then xc will divide the polynomial function resulting in a remainder of zero.

We will state this in the Factor Theorem.

Definition: FACTOR THEOREM

For any polynomial function f(x),

  • if xc is a factor of f(x), then f(c)=0
  • if f(c)=0, then xc is a factor of f(x)
Example 5.5.11

Use the Remainder Theorem to determine if x4 is a factor of f(x)=x364.

Solution

The Factor Theorem tells us that x4 is a factor of f(x)=x364 if f(4)=0.

f(x)=x364To evaluate f(4) substitute x=4.f(4)=4364Simplify.f(4)=6464Subtract.f(4)=0

Since f(4)=0,x4 is a factor of f(x)=x364.

Try It! 5.5.21

Use the Factor Theorem to determine if x5 is a factor of f(x)=x3125.

Answer

yes

Try It! 5.5.22

Use the Factor Theorem to determine if x6 is a factor of f(x)=x3216.

Answer

yes

Access these online resources for additional instruction and practice with dividing polynomials.

  • Dividing a Polynomial by a Binomial
  • Synthetic Division & Remainder Theorem

Key Concepts

  • Division of a Polynomial by a Monomial
    • To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
  • Division of Polynomial Functions
    • For functions f(x) and g(x), where g(x)0,
      (fg)(x)=f(x)g(x)
  • Remainder Theorem
    • If the polynomial function f(x) is divided by xc, then the remainder is f(c).
  • Factor Theorem: For any polynomial function f(x),
    • if xc is a factor of f(x), then f(c)=0
    • if f(c)=0, then xc is a factor of f(x)

This page titled 5.5: Dividing Polynomials is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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